AMC 10 · 2002 · #22
Grade 8 number-theoryPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Simulating all the way from 100 tile-by-tile is slow, so tool #5 (Look for a Pattern) leads: run the operation a couple of times and watch the count drop from 100. The key thing to notice is that only the number of tiles matters, because after each step the tiles are renumbered 1,2,3,… again. To see WHY the pattern holds, tool #4 (Introduce a Variable) lets us call the count n and note that one operation removes exactly the perfect squares ≤ n. Tool #9 (Solve an Easier Related Problem) sharpens this: test what happens when the count is a perfect square n=k², and a clean rule appears — two operations always turn k² tiles into (k-1)² tiles. Then counting the operations is just counting how many times you step a perfect square down from 10² to 1².
Run the operation twice and watch the count
Only the count matters: 100 tiles hold 10 squares, leaving 90; 90 tiles hold 9 squares, leaving 81=9².
Because the tiles are renumbered each time, the only thing that changes is the head count.
6.EE.A.1Look For A PatternCount removals with a square root
Call the count n. The squares up to n are 1²,…,⌊√(n)⌋², so each operation strips ⌊√(n)⌋ tiles: n → n-⌊√(n)⌋. At n=90 it strips 9 tiles.
The count of perfect squares up to n is just how far you can go before a square passes n, namely ⌊√(n)⌋.
The count of perfect squares up to a number is how far you can go before a square passes it.
▸ Why?
Every count of this kind is a whole part with a leftover, and only the whole part is counted.
▸ Why?
Squares only grow, so once one passes the limit every later square does too.
Two operations drop one perfect square
Try n=k²: dropping k leaves k²-k, just above (k-1)². The next drops k-1 and lands exactly on (k-1)²: two operations, one square down.
After pulling k tiles off a square count, you land just above the next lower square, so one more pull settles exactly onto it.
6.EE.A.3Solve An Easier Related ProblemStep the squares down and count
The base falls 10²→9²→…→1², and 1²=1 tile is the stop: 9 drops × 2 operations each = 18, choice (C).
Each pair of operations knocks the square base down by one, so nine knocks take 10² all the way to a single tile.
4.OA.C.5Look For A PatternOnly the tile count matters, and every two operations knock the perfect-square count down one step, so 10² tiles need 9 × 2=18 operations to reach a single tile.
- Run the operation twice and watch the count
- Count removals with a square root
- Two operations drop one perfect square
- Step the squares down and count